A. Field of the Invention
The present invention relates generally to orthogonal frequency division multiplexing (OFDM) systems used in digital wireless Communications systems and, more particularly to precoded OFDM systems robust to spectral null channels and vector OFDM systems with reduced cyclic prefix length.
B. Description of the Related Art
Orthogonal frequency division multiplexing (OFDM) systems have been widely used in high speed digital wireless communication systems, such as VHDSL and ADSL since OFDM systems convert intersymbol interference (ISI) channels into ISI-free channels by inserting a cyclic prefix as an overhead of the data rate at the transmitter. In high speed digital wireless applications, however, the ISI channel may have spectral nulls, which may degrade the performance of the existing OFDM systems because the Fourier transform of the ISI channel needs to be inverted for each subcarrier at the OFDM system receiver. For this reason, coded OFDM systems were proposed comprising conventional trellis coded modulation (TCM) or turbo codes. Another problem with conventional OFDM systems is that, when the ISI channel has many taps, the data rate overhead of the cyclic prefix insertion is high.
In a conventional OFDM system, as shown in FIG. 1, x(n) stands for the information symbol sequences after the binary to complex mapping, such as BPSK and QPSK symbol sequences, N is the number of carriers in the OFDM system, i.e., the size of the IFFT (inverted fast Fourier transform) and FFT (fast Fourier transform) in the OFDM system shown in FIG. 1 is N. The ISI channel has the following transfer function:                               H          ⁡                      (            z            )                          =                              ∑                          n              =              0                        L                    ⁢                                           ⁢                                    h              ⁡                              (                n                )                                      ⁢                          z                              -                n                                                                        (        2.1        )            where h(n) are the impulse responses of the ISI channel. Letting Γ be the cyclic prefix length in the OFDM system shown in FIG. 1 arid Γ>L for the purpose of removing the ISI; η(n) be the additive white Gaussian noise (AWGN) with mean zero and variance σ2=N0/2, where N0 is the single sided power spectral density of the noise η(n); and r(n) be the received signal at the receiver and y(n) be the signal after the FFT of the received signal r(n); then the relationship between the information symbols x(n) and the signal y(n) can be formulated as:yk(n)=Hkxk(n)+ξk(n), k=0, 1, . . . , N−1,  (2.2)where qk(n) denotes the kth subsequence of q(n), i.e., (q(n))n=(qo(n),q1(n), . . . , qN−1(n))n, and q stands for x, y, and ξ, ξ(n) is the FFT of the noise η(n) and therefore has the same statistics as η(n), andHk=H(z)|z=exp(j2πk/N), k=0, 1, . . . , N−1.  (2.3)The receiver needs to detect the information sequence xk(n) from yk(n) through Equation 2.2.
From Equation 2.2, the ISI channel H(z) is converted to N ISI-free subchannels Hk. The key for this property to hold is the inserting of the cyclic prefix with length Γ that is greater than or equal to the number of ISI taps L.
For the ISI-free system in Equation 2.2, the performance analysis of the detection is as follows: letting Pber,x(Eb/N0) be the bit error rate (BER) for the signal constellation x(n) in the AWGN channel at the SNR Eb/N0, where Eb is the energy per bit, then, the BER vs. Eb/N0 of the OFDM shown in FIG. 1 is:                               P          e                =                              1            N                    ⁢                                    ∑                              k                =                0                                            N                -                1                                      ⁢                                                   ⁢                                          P                                  ber                  ,                  x                                            ⁡                              (                                                                                                                                                      H                          k                                                                                            2                                        ⁢                                          NE                      b                                                                                                  (                                              N                        +                        Γ                                            )                                        ⁢                                          N                      0                                                                      )                                                                        (        2.4        )            For example, when the BPSK for x(n) is used, we have                                           P                          ber              ,              x                                ⁡                      (                                          E                b                            /                              N                0                                      )                          =                              Q            ⁡                          (                                                                    2                    ⁢                                          E                      b                                                                            N                    0                                                              )                                .                                    (        2.5        )            Therefore, the BER vs. Eb/N0 for the conventional OFDM system is                               P          e                =                              1            N                    ⁢                                    ∑                              k                =                0                                            N                -                1                                      ⁢                                                   ⁢                                          Q                ⁡                                  (                                                                                    2                        ⁢                                                                                                                                        H                              k                                                                                                            2                                                ⁢                                                  NE                          b                                                                                                                      (                                                      N                            +                            Γ                                                    )                                                ⁢                                                  N                          0                                                                                                      )                                            .                                                          (        2.6        )            